Introduction to the algorithm 10.4 a representation of a root tree

10.4-2 a two-fork tree given n nodes, writes out a recursive program of O (n) time, outputting the keywords for each node of the tree.

Pseudo code:

1 tree-print (T)2if T! = NIL3    PRINT key[t]4      tree-PRINT (left[t])5     tree-print (Right[t])

C Implementation:

void Treeprint (Tree t) {    if (t! = NULL)    {        printelement (t);        Treeprint (T-left);        Treeprint (T-right);    }}

10.4-3

(version one) pseudo-code:

1 tree-PRINT (T, S)2 while    t! = Nil or S! = Nil3       if T! = NIL4          PUSH (S, T)5          VISIT (t)6           T = t->left7       Else 8           t = POP (S)9           t = t->right

(version II) pseudo-code:

1tree-PRINT (T, S)2   ifT! =NIL3 PUSH (T, S)4        whileS! =NIL5Node =POP (S)6 VISIT (Node)7           ifLeft[node]! =NIL8 PUSH (Left[node], S)9           ifRight[node]! =NILTenPUSH (Right[node], S)

10.4-4

Analysis: The root tree is stored in a two-tree form and can be processed using the above method

10.4-5

Analysis: Adds a flag to the node and a pointer to the parent, which is the storage space of the tree itself, which uses a fixed amount of additional storage space

Pseudo code:

1 treeprint (T)2    whileT! =NIL3        whileLEFT[T]! = NIL and left[t]. Visited! =true4T =Left[t]5       ift.visited =false6 VISIT (T)7t.visited =true8       ifRIGHT[T]! = NIL and right[t]. Visited! =true9T =Right[t]Ten       Else OneT = Parent[t]

10.4-6

Solution: For a root tree, using binary tree storage, each node contains a left and right pointers and a bool value;

Set each node in the root tree, and if it has a child, set its last child's pointer to its father, which is the node, and set bool to true;

If there is a root tree node, the child's right pointer points to the sibling node, set the bool value to false;

Introduction to the algorithm 10.4 a representation of a root tree